Question

[Trần Minh Hiển]

       Cho x,y,z>0 C/M

\(\sqrt{\dfrac{x^3}{y^3}}+\sqrt{\dfrac{y^3}{z^3}}+\sqrt{\dfrac{z^3}{x^3}}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\)

Answer 1

[Nguyễn Việt Lâm]

\(\sqrt{\dfrac{x^3}{y^3}}+\sqrt{\dfrac{x^3}{y^3}}+1\ge\dfrac{3x}{y}\) ; \(2\sqrt{\dfrac{y^3}{z^3}}+1\ge\dfrac{3y}{z}\) ; \(2\sqrt{\dfrac{z^3}{x^3}}+1\ge\dfrac{3z}{x}\)

\(\Rightarrow2VT+3\ge\dfrac{3x}{y}+\dfrac{3y}{z}+\dfrac{3z}{x}\)

\(\Rightarrow2VT+3\ge\dfrac{2x}{y}+\dfrac{2y}{z}+\dfrac{2z}{x}+3\sqrt[3]{\dfrac{xyz}{xyz}}\)

\(\Rightarrow VT\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\) (đpcm)